Answer
$16-16i$
Work Step by Step
$-2-2i$ in trigonometric form is
$-2-2i=2\sqrt {2}(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(-2-2i)^{3}=[2\sqrt {2}(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4})]^{3}=(2\sqrt {2})^{3}(\cos 3\cdot\frac{5\pi}{4}+i\sin 3\cdot\frac{5\pi}{4})$
$=16\sqrt {2}(\cos \frac{15\pi}{4}+i\sin\frac{15\pi}{4})$
$=16\sqrt {2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$
($\frac{7\pi}{4}$ and $\frac{15\pi}{4}$ are coterminal.)
In standard form, our result is
$=16\sqrt {2}(\frac{\sqrt 2}{2}+i\cdot-\frac{\sqrt 2}{2})$
$=16-16i$
